- A$\frac{1}{\text{RC}}$
- B$\frac{\text{R}}{\text{L}}$
- C$\frac{1}{\sqrt{\text{LC}}}$
- D$\text{C}/\text{L}$
$\frac{1}{\text{RC}}$
$\frac{\text{R}}{\text{L}}$
$\frac{1}{\sqrt{\text{LC}}}$
Explanation:
þ Time constant t = RC in RC circuit
frequency $=\frac{1}{\tau}=\frac{1}{\text{RC}} \ ...(\text{i})$
þ Time constant in LR circuit is $\tau=\frac{\text{L}}{\text{RC}}$
frequency $\frac{1}{\tau}=\frac{\text{R}}{\text{L}} \ ...(\text{ii})$
þ eq. (i) & (ii) multiply
frequency $=\frac{1}{\text{LC}}$
frequency $=\frac{1}{\sqrt{2\text{C}}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
A proton and an α - particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 μ sec to make 5 revolutions, then the periodic time for the α - particle would be
|
(a) 50 μ sec |
(b) 25 μ sec |
(c) 10 μ sec |
(d) 5 μ sec |

A transformer has turn ratio 100/1. If secondary coil has 4 amp current then current in primary coil is
|
(a) 4 A |
(b) 0.04 A |
(c) 0.4 A |
(d) 400 A |

Correct exposure for a photographic print is 10 seconds at a distance of one metre from a point source of 20 candela. For an equal fogging of the print placed at a distance of 2 m from a 16 candela source, the necessary time for exposure is
|
(a) 100 sec |
(b) 25 sec |
(c) 50 sec |
(d) 75 sec |
Force acting upon a charged particle kept between the plates of a charged condenser is F. If one plate of the condenser is removed, then the force acting on the same particle will become
|
(a) 0 |
(b) F/2 |
(c) F |
(d) 2F |
Figure here shows P and Q as two equally intense coherent sources emitting radiations of wavelength 20 m. The separation PQ is 5.0 m and phase of P is ahead of the phase of Q by 90o. A, B and C are three distant points of observation equidistant from the mid-point of PQ. The intensity of radiations at A, B, C will bear the ratio
|
(a) 0 : 1 : 4 |
(b) 4 : 1 : 0 |
(c) 0 : 1 : 2 |
(d) 2 : 1 : 0 |